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Here you will learn about the quadratic formula and how you can use it to solve quadratic equations.
Students will first learn about the quadratic formula as part of algebra in high school.
The quadratic formula is used to solve quadratic equations by finding the roots, x.
The quadratic formula is: x=\cfrac{-b\pm\sqrt{b^2-4ac}}{2a}
By using the general form of a quadratic equation, a x^{2}+b x+c=0, you can substitute the values of a, b and c into the quadratic formula to calculate x (the solution(s) for the quadratic formula).
The \pm in front of the square root means βplus or minus.β This accounts for the positive root and negative root.
Quadratic equations normally have two solutions, so you need to use the formula twice, once with a + and once with a β.
x=\cfrac{-b+\sqrt{b^2-4ac}}{2a}\quad x=\cfrac{-b-\sqrt{b^2-4ac}}{2a}
To find out how many real roots a quadratic equation has, use the expression under the square root in the quadratic formula β b^2-4 a c. This is called the discriminant.
For example,
x^2+3x+2=0Letβs see how many real solutions this quadratic equation has.
\begin{aligned} & b^2-4 ac \\\\ & =3^2-4 \times 1 \times 2 \\\\ & =9-8 \\\\ & =1 \end{aligned}Because the discriminant is greater than 0, there are two solutions (real numbers).
Now, letβs solve for the two solutions.
\begin{aligned} & x=\cfrac{-3+\sqrt{3^2-4 \times 1 \times 2}}{2 \times 1} \\\\ & x=\cfrac{-3+\sqrt{1}}{2 \times 1} \\\\ & x=\cfrac{-3+1}{2} \\\\ & x=\cfrac{-2}{2} \\\\ & x=-1 \end{aligned} \quad \begin{aligned} & x=\cfrac{-3-\sqrt{3^2-4 \times 1 \times 2}}{2 \times 1} \\\\ & x=\cfrac{-3-\sqrt{1}}{2 \times 1} \\\\ & x=\cfrac{-3-1}{2} \\\\ & x=\cfrac{-4}{2} \\\\ & x=-2 \end{aligned}
You can check by graphing the solution. Remember, the solutions are the x -intercepts.
How does this relate to high school math?
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DOWNLOAD FREEIn order to use the quadratic formula:
Solve x^{2}-8x+15=0.
2Substitute these values into the quadratic formula.
\begin{aligned} & x=\cfrac{-b\pm\sqrt{b^2-4ac}}{2a} \\\\ & x=\cfrac{-(-8)\pm\sqrt{(-8)^2-4(1)(15)}}{2(1)} \end{aligned}Using parentheses will help to make the calculation clear.
3Solve the equation with a \bf{+}, and then with a \bf{-}.
\begin{aligned} & x=\cfrac{-(-8)+\sqrt{(-8)^2-4(1)(15)}}{2(1)} \\\\ & x=\cfrac{8+\sqrt{64-60}}{2} \\\\ & x=\cfrac{8+2}{2} \\\\ & x=5 \end{aligned}\quad \begin{aligned} & x=\cfrac{-(-8)-\sqrt{(-8)^2-4(1)(15)}}{2(1)} \\\\ & x=\cfrac{8-\sqrt{64-60}}{2} \\\\ & x=\cfrac{8-2}{2} \\\\ & x=3\end{aligned}
When you plot the graph of the quadratic equation you get a special βUβ shaped curve called a parabola.
By using graphing techniques such as this, you can see that the roots of the quadratic formula are where the quadratic graph crosses the \textbf{x} -axis or the x -intercepts.
You also can check that the solutions are correct by substituting them back into the quadratic equation.
Solve 2x^{2}+5x+3=0.
Identify the value of \textbf{a, b} and \textbf{c} in a quadratic equation.
Substitute these values into the quadratic formula.
Solve the equation with a \bf {+}, and then with a \bf {-}.
\begin{aligned} & x=\cfrac{-(5)+\sqrt{(5)^2-4(2)(3)}}{2(2)} \\\\ & x=\cfrac{-5+\sqrt{25-24}}{4} \\\\ & x=\cfrac{-5+1}{4} \\\\ & x=-1 \end{aligned}\quad \begin{aligned} & x=\cfrac{-(5)-\sqrt{(5)^2-4(2)(3)}}{2(2)} \\\\ & x=\cfrac{-5-\sqrt{25-24}}{4} \\\\ & x=\cfrac{-5-1}{4} \\\\ & x=-1.5\end{aligned}
You can see the roots of the quadratic equation are where the quadratic graph crosses the x -axis.
You can also check that the solutions are correct by substituting them back into the quadratic equation.
Solve -6x^2+x+2=0.
Identify the value of \textbf{a, b} and \textbf{c} in a quadratic equation.
Substitute these values into the quadratic formula.
Solve the equation with a \bf {+}, and then with a \bf {-}.
\begin{aligned} & x=\cfrac{-(1)+\sqrt{(1)^2-4(-6)(2)}}{2(-6)} \\\\ & x=\cfrac{-1+\sqrt{1-(-48)}}{-12} \\\\ & x=\cfrac{-1+7}{-12} \\\\ & x=-\cfrac{1}{2} \end{aligned} \quad \begin{aligned} & x=\cfrac{-(1)-\sqrt{(1)^2-4(-6)(2)}}{2(-6)} \\\\ & x=\cfrac{-1-\sqrt{1-(-48)}}{-12} \\\\ & x=\cfrac{-1-7}{-12} \\\\ & x=\cfrac{2}{3}\end{aligned}
You can see the roots of the quadratic equation are where the quadratic graph crosses the x -axis.
You can also check that the solutions are correct by substituting them back into the quadratic equation.
Solve x^2+14=9 x.
Identify the value of \textbf{a, b} and \textbf{c} in a quadratic equation.
In order to solve the following equation, x^2+14=9x, all terms of the equation need to be on the left-hand side of the equation.
\begin{aligned}
& x^2+14=9 x \\\\
& -9 x-9 x \\\\
& x^2-9 x+14=0
\end{aligned}
\begin{aligned}
& a=1 \\\\
& b=-9 \\\\
& c=14
\end{aligned}
Substitute these values into the quadratic formula.
Solve the equation with a \bf {+}, and then with a \bf {-}.
\begin{aligned} & x=\cfrac{-\left(-9)+\sqrt{(-9)^2-4(1)(14)}\right.}{2(1)} \\\\ & x=\cfrac{9+\sqrt{81-56}}{2} \\\\ & x=\cfrac{9+5}{2} \\\\ & x=7 \end{aligned} \quad \begin{aligned} & x=\cfrac{-(-9)-\sqrt{(-9)^2-4(1)(14)}}{2(1)} \\\\ & x=\cfrac{9-\sqrt{81-56}}{2} \\\\ & x=\cfrac{9-5}{2} \\\\ & x=2\end{aligned}
You can see the roots of the quadratic equation are where the quadratic graph crosses the x -axis.
You can also check that the solutions are correct by substituting them back into the quadratic equation.
Solve x^2-x+1=0.
Identify the value of \textbf{a, b} and \textbf{c} in a quadratic equation.
Substitute these values into the quadratic formula.
Solve the equation with a \bf {+}, and then with a \bf {-}.
\begin{aligned} & x=\cfrac{-(-1)+\sqrt{(-1)^2-4(1)(1)}}{2(1)} \\\\ & x=\cfrac{1+\sqrt{1-4}}{2} \\\\ & x=\cfrac{1+\sqrt{3} i}{2} \\\\ & x=0.5+0.866 i \end{aligned} \quad \begin{aligned} & x=\cfrac{-(-1)-\sqrt{(-1)^2-4(1)(1)}}{2(1)} \\\\ & x=\cfrac{1-\sqrt{1-4}}{2} \\\\ & x=\cfrac{1-\sqrt{3} i}{2} \\\\ & x=0.5-0.866 i \end{aligned}
Note, since \sqrt{1-4}=\sqrt{-3}, you use a complex number to represent the square root.
The roots are imaginary numbers, because this parabola does not cross the x -axis.
Solve 2x^2-7 x+19=0.
Identify the value of \textbf{a, b} and \textbf{c} in a quadratic equation.
Substitute these values into the quadratic formula.
Solve the equation with a \bf{+}, and then with a \bf{-}.
\begin{aligned} & x=\cfrac{-(-7)+\sqrt{(-7)^2-4(2)(19)}}{2(2)} \\\\ & x=\cfrac{7+\sqrt{49-152}}{4} \\\\\ & x=\cfrac{7+\sqrt{103 i}}{4} \\\\ & x=1.75+2.537 i \end{aligned} \quad \begin{aligned} & x=\cfrac{-(-7)-\sqrt{(-7)^2-4(2)(19)}}{2(2)} \\\\ & x=\cfrac{7-\sqrt{49-152}}{4} \\\\ & x=\cfrac{7-\sqrt{103 i}}{4} \\\\ & x=1.75-2.537 i \end{aligned}
Note, since \sqrt{49-152}=\sqrt{-103}, you use a complex number to represent the square root.
The roots are imaginary numbers, because this parabola does not cross the x -axis.
x^{2}-8x+15=0
so here b=-8, NOT 8 or -8x 2x^{2}+5x+3=0,
a=2,\qquad b=5,\qquad c=3
x^{2}-4 x+5=0
1) Solve {x}^2+4x+3=0 using the quadratic formula.
Identify a, b and c.
a=1, \quad b=4, \quad c=3
Substitute these values into the quadratic formula.
\begin{aligned} & x=\cfrac{-b \pm \sqrt{b^2-4 a c}}{2 a} \\\\ & x=\cfrac{-4 \pm \sqrt{(4)^2-4(1)(3)}}{2(1)} \end{aligned}
Solve the equation with a +, and then with a -.
\begin{aligned} & x=\cfrac{-4+\sqrt{(4)^2-4(1)(3)}}{2(1)} \\\\ & x=\cfrac{-4+\sqrt{16-12}}{2} \\\\ & x=\cfrac{-4+2}{2} \\\\ & x=-1 \end{aligned} \quad \begin{aligned} & x=\cfrac{-4-\sqrt{(4)^2-4(1)(3)}}{2(1)} \\\\ & x=\cfrac{-4-\sqrt{16-12}}{2} \\\\ & x=\cfrac{-4-2}{2} \\\\ & x=-3 \end{aligned}
2) Solve {x}^2+3x-18=0 using the quadratic formula.
Identify a, b and c.
a=1, \quad b=3, \quad c=-18
Substitute these values into the quadratic formula.
\begin{aligned}
& x=\cfrac{-b \pm \sqrt{b^2-4 a c}}{2 a} \\\\
& x=\cfrac{-3 \pm \sqrt{(3)^2-4(1)(-18)}}{2(1)}
\end{aligned}
Solve the equation with a +, and then with a -.
\begin{aligned}
& x=\cfrac{-3+\sqrt{(3)^2-4(1)(-18)}}{2(1)} \\\\
& x=\cfrac{-3+\sqrt{9-(-72)}}{2} \\\\
& x=\cfrac{-3+9}{2} \\\\
& x=3 \end{aligned} \quad \begin{aligned}
& x=\cfrac{-3-\sqrt{(3)^2-4(1)(-18)}}{2(1)} \\\\
& x=\cfrac{-3-\sqrt{9-(-72)}}{2} \\\\
& x=\cfrac{-3-9}{2} \\\\
& x=-6
\end{aligned}
3) Solve 2{x}^2-3x-2=0 using the quadratic formula.
Identify a, b and c.
a=2, \quad b=-3, \quad c=-2
Substitute these values into the quadratic formula.
\begin{aligned}
& x=\cfrac{-b \pm \sqrt{b^2-4 a c}}{2 a} \\\\
& x=\cfrac{-(-3) \pm \sqrt{(-3)^2-4(2)(-2)}}{2(2)}
\end{aligned}
Solve the equation with a +, and then with a -.
\begin{aligned}
& x=\cfrac{-(-3)+\sqrt{(-3)^2-4(2)(-2)}}{2(2)} \\\\
& x=\cfrac{3+\sqrt{9-(-16)}}{4} \\\\
& x=\cfrac{3+5}{4} \\\\
& x=2
\end{aligned} \quad \begin{aligned}
& x=\cfrac{-(-3)-\sqrt{(-3)^2-4(2)(-2)}}{2(2)} \\\\
& x=\cfrac{3-\sqrt{9-(-16)}}{4} \\\\
& x=\cfrac{3-5}{4} \\\\
& x=-0.5
\end{aligned}
4) Solve -x^2+2 x+1=-11 using the quadratic formula.
First, convert the equation to the general form ax^2+b x+c=0.
\begin{aligned}
-x^2+2 x+1= & -11 \\\\
+11 \hspace{0.2cm} & +11 \\\\
-x^2+2 x+12= & 0
\end{aligned}
Identify a, b and c.
a=-1, \quad b=2, \quad c=12
Substitute these values into the quadratic formula.
\begin{aligned}
& x=\cfrac{-b \pm \sqrt{b^2-4 a c}}{2 a} \\\\
& x=\cfrac{-2 \pm \sqrt{(2)^2-4(-1)(12)}}{2(-1)}
\end{aligned}
Solve the equation with a +, and then with a -.
\begin{aligned}
& x=\cfrac{-2+\sqrt{(2)^2-4(-1)(12)}}{2(-1)} \\\\
& x=\cfrac{-2+\sqrt{4-(-48)}}{-2} \\\\
& x=\cfrac{-2+7.211}{-2} \\\\
& x=-2.6055
\end{aligned} \quad \begin{aligned}
& x=\cfrac{-2-\sqrt{(2)^2-4(-1)(12)}}{2(-1)} \\\\
& x=\cfrac{-2-\sqrt{4-(-48)}}{-2} \\\\
& x=\cfrac{-2-7.211}{-2} \\\\
& x=4.6055
\end{aligned}
5) Solve -x^2+4x-20=0 using the quadratic formula.
Identify a, b and c.
a=-1, \quad b=4, \quad c=-20
Substitute these values into the quadratic formula.
\begin{aligned}
& x=\cfrac{-b \pm \sqrt{b^2-4 a c}}{2 a} \\\\
& x=\cfrac{-4 \pm \sqrt{(4)^2-4(-1)(-20)}}{2(-1)}
\end{aligned}
Solve the equation with a +, and then with a -.
\begin{aligned}
& x=\cfrac{-4+\sqrt{(4)^2-4(-1)(-20)}}{2(-1)} \\\\
& x=\cfrac{-4+\sqrt{16-80}}{-2} \\\\
& x=\cfrac{-4+\sqrt{64} i}{-2} \\\\
& x=2-4 i \end{aligned} \quad \begin{aligned}
& x=\cfrac{-4-\sqrt{(4)^2-4(-1)(-20)}}{2(-1)} \\\\
& x=\cfrac{-4-\sqrt{16-80}}{-2} \\\\
& x=\cfrac{-4-\sqrt{64} i}{-2} \\\\
& x=2+4 i
\end{aligned}
You can see that there are no values of x that give a y value of 0 .
The graph does not cross the x-axis. The use of the quadratic formula will result in roots that are not real.
6) Solve 3x^2-7x+21=0 using the quadratic formula.
Identify a, b and c.
a=3, \quad b=-7, \quad c=21
Substitute these values into the quadratic formula.
\begin{aligned}
& x=\cfrac{-b \pm \sqrt{b^2-4 a c}}{2 a} \\\\
& x=\cfrac{-(-7) \pm \sqrt{(-7)^2-4(3)(21)}}{2(3)}
\end{aligned}Β
Solve the equation with a +, and then with a -.
\begin{aligned}
& x=\cfrac{-(-7)+\sqrt{(-7)^2-4(3)(21)}}{2(3)} \\\\
& x=\cfrac{7+\sqrt{49-252}}{6} \\\\
& x=\cfrac{7+\sqrt{203 i}}{6} \\\\
& x=1.1 \overline{6}+2.3746 i
\end{aligned}Β \quad \begin{aligned}
& x=\cfrac{-(-7)-\sqrt{(-7)^2-4(3)(21)}}{2(3)} \\\\
& x=\cfrac{7-\sqrt{49-252}}{6} \\\\
& x=\cfrac{7-\sqrt{203} i}{6} \\\\
& x=1.1 \overline{6}-2.3746 i
\end{aligned}Β
You can see that there are no values of x that give a y value of 0.
The graph does not cross the x -axis. The use of the quadratic formula will result in roots that are not real.
No, it only works for algebraic equations that have a second-degree term, otherwise known as quadratic equations. It will not work for polynomials of degrees other than two, such as linear (one-degree) or cubic equations (three-degree).
When the minimum (vertex) of the parabola formed by the quadratic equation is on the x -axis, it will only have one solution.
Yes, a can be positive or negative in a quadratic expression or equation. A negative a value causes the graph of the parabola to be flipped and open downward, instead of upward.
No, mathematicians have developed various ways to solve for the roots of a quadratic equation. Other ways include factoring the quadratic into the product of two terms or utilizing perfect squares to solve with a method called completing the square.
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